This semester in my differential geometry class the professor stated the following without further comment:
Let $N$ be a compact smooth manifold in $\mathbf{R}^n$. For $\varepsilon > 0$ small enough, there exists a canonical retraction $\rho$ from the $\varepsilon$-neighborhood $U$ of $N$ to $N$, which is a submersion.
I thought that $\rho(x)$ might be defined as the closest point of $N$ to $x$, which is well-defined, as $N$ is compact. Then $\rho$ is a continuous map and thus clearly a retraction onto $N$. But why should $\rho$ be a submersion?
Thanks!
Your idea is basically correct, but I think things get much easier if you turn things around and start from the normal bundle of $N$. So take $E:=\{(x,v)\in N\times\mathbb R^n:v\perp T_xN\}$ and the map $\phi:E\to\mathbb R^n$ defined by $\phi(x,v):=x+v$. This is immediately seen to have invertible differential in $(x,0)$ for all $x\in N$ and hence is a diffeomorphism locally around each of these points. Compactness of $N$ easily implies that there is an $\epsilon>0$ such that $\phi$ restricts to a diffeomorphism from $W:=\{(x,v)\in E:\|v\|<\epsilon\}$ onto an open neighborhood $U$ of $N$ in $\mathbb R^n$. Then you can define $\rho:U\to N$ as the composition of the first projection $W\to N$ (which clearly is a surjective submersion) with $\phi^{-1}$.